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Air resistance of baseball

  1. Sep 18, 2007 #1
    When a baseball flies through the air, the ratio f[tex]_{quad}[/tex] / f [tex]_{lin}[/tex] of the quadratic to the linear drag force is given by

    [tex]\frac{f_{quad}}{f_{lin}}[/tex] = [tex]\frac{cv^{2}}{bv}[/tex] = [tex]\frac{\gamma D}{\beta}[/tex] v = (1.6 x 10[tex]^{3}[/tex] [tex]\frac{s}{m^{2}}[/tex]) Dv.

    Given that a baseball has a diamater of 7 cm, find the approximate speed v at which the two drag forces are equally important. For what approximate range of speeds is it sage to treat the drag foce as purely quadratic? Under normal conditions is it a good approximation to ignore the linear term?



    f[tex]_{lin}[/tex] = bv

    f[tex]^{quad}[/tex] = cv[tex]^{2}[/tex]



    Dumb question in starting this, does the b here represent slope? I can't find a definition of the variable in my text....





    3. The attempt at a solution
     
  2. jcsd
  3. Sep 18, 2007 #2
    I found that

    b = [tex]\beta[/tex]D

    c = [tex]\gamma[/tex]D[tex]^{2}[/tex]

    I'm not sure how the ratio works so that you find which force is negligable and which is not.
    Help?
     
  4. Sep 18, 2007 #3

    learningphysics

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    For the first part, at what velocity in terms of b and c, is the quadratic component equal to the linear component ?

    my guess is, it's safe to ignore the linear term if the quadratic force is ten times the linear force or more.
     
  5. Sep 18, 2007 #4
    How does that equation though, since the two thing being compared and being divided, show what can be ignored?
     
  6. Sep 18, 2007 #5

    learningphysics

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    It's showing you the ratio of the two forces... if the ratio is high then the linear term is insignificant compared to the quadratic term (since the quadratic term is so much bigger)... so the linear term won't have much of an effect as compared to the quadratic term... so we might as well ignore it.

    On the other hand if the ratio is extremely low... close to 0, then the quadratic term is insignificant compared to the linear term... and we can ignore the quadratic term, and keep the linear term.
     
  7. Sep 18, 2007 #6

    learningphysics

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    I used a ratio of 10, purely as a guess... I don't know what would be a good ratio... 10 times seems big enough...
     
  8. Sep 18, 2007 #7
    I see.....
    So, i want it equal to 1 for them to be equally important?
     
  9. Sep 18, 2007 #8

    learningphysics

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    yes, exactly.
     
  10. Sep 18, 2007 #9
    Ok but the velocity is always squared in the quadratic force. How can I ever change the velocity to make them equal?
     
  11. Sep 18, 2007 #10

    learningphysics

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    Solve for the velocity. You'll see.
     
  12. Sep 18, 2007 #11
    I got 9.14 x 10^-3 m/s.... that doesn't sound right.
     
  13. Sep 18, 2007 #12
    the values given for beta is 1.6 x 10^-4 Ns/m^2

    and gamma is 0.25 Ns^2/m^4
     
  14. Sep 18, 2007 #13

    learningphysics

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    What is b and c?
     
  15. Sep 18, 2007 #14

    learningphysics

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    You want 1.6*10^3*D*v = 1. using D= 0.07 I get 8.93*10^-3m/s
     
  16. Sep 18, 2007 #15
    b: I get (1.6 x 10^-4) x (7cm) = 11.2 x 10^-6

    c: (.25) x (7cm^2) = 1.225 x 10^-3
     
  17. Sep 18, 2007 #16
    It matters what part of the equation you use?
     
  18. Sep 18, 2007 #17

    learningphysics

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    No. It's all the same. But it's probably most convenient to use 1.6*10^3 D*v, since you can just plug in the diameter.
     
  19. Sep 18, 2007 #18
    I must be doing something wrong with the other part...
     
  20. Sep 18, 2007 #19
    "For what approximate range of speeds is it safe to treat the drag force as purely quadratic".

    ie when is linear negliable?
     
  21. Sep 18, 2007 #20

    learningphysics

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    Not sure... it's a matter of opinion. When the ratio is 10 or more I'd say...

    I think the reason our numbers are off are because gamma/beta = 1.5625*10^3... not 1.6*10^3

    when I use 1.5625*10^3, I also get 9.14*10^-3m/s. So I think that's right.
     
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